Suppose $m$ is an integer such that $9m^2 + 25m + 26$ is the product of two consecutive integers. Find $m.$
I first let $k$ be equal to the larger of the two consecutive integers so that I can set up the equation $9m^2 + 25m + 26 = k(k-1).$ However, I am unsure where to go from here. Can someone give me a hint please?
On
One possible way :
Let $M=9m^2 + 25m + 26$
Then discriminant, $D$ of $k^2 \pm k - M=0$ is a perfect square for integer root(s) $k$. So $D = 1 + 4M$ is a perfect square.
$$4M + 1 = 36m^2 + 100m + 105$$
Compare with nearby squares $$ \begin{align} (6m + 8)^2 \cdots 36m^2 + 100m + 105 \cdots (6m + 9)^2 \\ 36m^2 + 96m + 64 < 36m^2 + 100m + 105 \cdots 36m^2 + 108m + 81 \\ \end{align} $$
For $$ 36m^2 + 100m + 105 = 36m^2 + 108m + 81 $$
$m=3$ and $k=14$, $-13$.
The key insight is completing the squares.
Let $y=18m+25$. Then $9m^2 + 25m + 26 = k(k-1)$ becomes $y^2+311=(6k-3)^2-9$. So, it boils down to writing $320$ as difference of two squares, which is easy given the factorization of $320$. There are only seven solutions of $320=ab$ with $0\le a\le b$. Only one works and gives $m=3$ and $k=14$. If you allow negative numbers, then there is also $m=3$ and $k=-13$, $m=-2$ and $k=-3$, $m=-2$ and $k=4$.